The diagonal of a rectangle table is 9 feet, and the length of the table is 5 feet more than its width. Find the length and width of the table.

Question

The diagonal of a rectangle table is 9 feet, and the length of the table is 5 feet more than its width. Find the length and width of the table.

Answer 1

the length, width, and diagonal form a right triangle

w=width

w+5=length

9=diagonal (hypotenuse)

9^2=w^2+(w+5)^2

81=w^2+w^2+10w+25

w^2+w^2+10w+25-81=0

2w^2+10w-56=0

w^2+5w-28=0

factor using quadratic formula

[-5±√(25+112)]/2

[-5±√137]/2

[-5±11.7]/2

6.7/2

3.35 feet=w, the width

3.35+5=8.35 feet=length

check: 3.35^2+8.35^2=9^2

          11.2225+69.7225=81

          80.945≈81

Answer 2

Let w = width; length is w + 5. The formula for diagonal is the square root of the length squared plus the width squared, so we have: d = 9 feet = SQRT (w^2 + [w + 5]^2). Squaring both sides, we have:

81 = w^2 + w^2 + 10 w + 25, or 2w^2 + 10w - 56 = 0. Solving with solution to quadratic equation, we get width of 3.35 feet and length of 8.35 feet. Check the answer by adding squared of length and width, then taking the square root of that sum.